Life In 19x19

RobertJasiek wrote:But where in your scheme do P ≥ Q and P ≤ Q fit?

It's not my scheme, it's Conway, Berlekamp, and Guy's scheme.

Let Black play first in G, with correct play. The results are the opposite if White plays first.

1) If Black wins, then G |> 0.

2) If White wins, then G ≤ 0.

Bill Spight wrote: Let Black play first in G, with correct play. The results are the opposite if White plays first.

1) If Black wins, then G |> 0.

2) If White wins, then G ≤ 0.

I think you mean

2) If White wins or ties (that is, Black cannot win), then G ≤ 0.

RobertJasiek wrote:

Bill Spight wrote: Let Black play first in G, with correct play. The results are the opposite if White plays first.

1) If Black wins, then G |> 0.

2) If White wins, then G ≤ 0.

I think you mean

2) If White wins or ties (that is, Black cannot win), then G ≤ 0.

No, I meant in the CGT sense. Either player can win a tie.

I came up with the "cannot win" formulation for certain difference games, such as those for comparing two options or for testing for reversal. The point being that in the difference game P - Q for those questions, P - Q has the same number of stones of each color and an even number of empty points. If you get a jigo result it should come after a even number of plays and thus indicate a second player win in CGT. That has been my experience. But that is not the case in general.

I did not want to teach unfamiliar CGT concepts for people to be able to use difference games for those purposes. However, it is certainly possible for jigo to occur in such difference games with an odd number of plays and an odd number of dame, but, IMX, that would be a rare occurrence. I think I was wrong not to address that issue. You can do so in terms of sente and gote, I think, without having to invoke CGT. And, IMX, it is very unlikely to come up, anyway.

Bill Spight wrote: What you should be interested in is scores, not counts. If you do not play when the position has a score, then in {-1 | 1} White does not play to 1, but stops at 0; i.e., does not play at all. That solves that problem.

The particular scores you are interested in are stops. A stop is the first score reached with optimal minimax play. For game, G, let S(B) be the stop when Black plays first and S(W) be the stop when White plays first. OC, S(B) ≥ S(W).

1) If S(B) < 0 then G < 0.

2) If S(W) > 0 then G > 0.

3) If S(B) > 0 > S(W) then G <> 0.

4) If S(B) = 0 and S(W) < 0 then G <| 0.

5) If S(B) > 0 and S(W) = 0 then G |> 0.

6) If S(B) = S(W) = 0, then G = 0 or G is an infinitesimal.

Now, we may not always realize when we have reached a score without playing the game out. For example, {3|3||0|||0||-5|-7} = 0.

Also, for difference games who gets the last play in a jigo may well matter, because sente usually matters.

In in the great majority of cases when S(B) = S(W) = 0 the optimal sequences are gote for who plays first.
Do you have an example in which one of the optimal sequence (black plays first or white plays first) is sente with S(B) = S(W) = 0 ?

Gérard TAILLE wrote:In in the great majority of cases when S(B) = S(W) = 0 the optimal sequences are gote for who plays first.

Not IMX. The positive and negative infinitesimals are sente for one side, gote for the other. Then you have infinitesimals that are confused with 0, which are gote for both sides, and the cases where G = 0, which are sente for both sides, unless a play by either side is non-existent or a loss.

Gérard TAILLE wrote:Do you have an example in which one of the optimal sequence (black plays first or white plays first) is sente with S(B) = S(W) = 0 ?

Well, Robert's example, where G = 0, is one.

Click Here To Show Diagram Code

[go]$$B Example 1: P + (-Q)
$$ ---------------------------------
$$ | . . . . . . O . X . . X O . . |
$$ | X X X X O O O . X X X O O O O |
$$ | . . . . . . . . . . . . . . . |[/go]

P - Q has the same number of Black stones and White stones, and an even number of empty points. A jigo result is thus very likely to be sente for both sides.

Bill Spight wrote:

Gérard TAILLE wrote:In in the great majority of cases when S(B) = S(W) = 0 the optimal sequences are gote for who plays first.

Not IMX. The positive and negative infinitesimals are sente for one side, gote for the other. Then you have infinitesimals that are confused with 0, which are gote for both sides, and the cases where G = 0, which are sente for both sides, unless a play by either side is non-existent or a loss.

Gérard TAILLE wrote:Do you have an example in which one of the optimal sequence (black plays first or white plays first) is sente with S(B) = S(W) = 0 ?

Well, Robert's example, where G = 0, is one.

$$B Example 1: P + (-Q)
$$ ---------------------------------
$$ | . . . . . . O . X . . X O . . |
$$ | X X X X O O O . X X X O O O O |
$$ | . . . . . . . . . . . . . . . |

Click Here To Show Diagram Code

[go]$$B Example 1: P + (-Q)
$$ ---------------------------------
$$ | . . . . . . O . X . . X O . . |
$$ | X X X X O O O . X X X O O O O |
$$ | . . . . . . . . . . . . . . . |[/go]

P - Q has the same number of Black stones and White stones, and an even number of empty points. A jigo result is thus very likely to be sente for both sides.

Oops I wrote the contrary of what I had in mind. Sorry for that.
In the majority of cases, when S(B) = S(W) = 0 in a difference game, the two areas P and -Q are miai and the sequence is sente for who plays first (=> in CGT context it is a loss for the player who plays first)
Do you know situations where one of the sequence is gote and S(B) = S(W) = 0?

Gérard TAILLE wrote:In the majority of cases, when S(B) = S(W) = 0 in a difference game, the two areas P and -Q are miai and the sequence is sente for who plays first (=> in CGT context it is a loss for the player who plays first)
Do you know situations where one of the sequence is gote and S(B) = S(W) = 0?

How silly of me! Consider {0,*|0}. If we were to use a difference game to choose between the Black options, we could get DG = * - 0 = *, which is confused with 0.

OC, such examples are not uncommon in chilled go, but I don't remember finding one in regular go when I was asking that question. But all you need is a choice between two seki with different parity.

In the video https://www.youtube.com/watch?v=3kj3YQb9P74 Michael Redmond propose the following postion :

Click Here To Show Diagram Code

[go]$$B White to play
$$ -----------------------------------------
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . O . . . O . O . . . . |
$$ | . . . . . . . . . . . . . . O X X . . |
$$ | . . . . . . . . . . . . . . . O . . . |
$$ | . . . . . . . . . . . . . . . O O . . |
$$ | . . . . . . . . . . . . . . X O X . . |
$$ | . . . . . . . . . . . . . . . X X . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ -----------------------------------------[/go]

and shows the failure of white in the sequence:

Click Here To Show Diagram Code

[go]$$B White to play
$$ -----------------------------------------
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . O . . . O . O . . . . |
$$ | . . . . . . . . . . . . . . O X X . . |
$$ | . . . . . . . . . . . . . . 7 O 3 2 6 |
$$ | . . . . . . . . . . . . . . . O O 1 5 |
$$ | . . . . . . . . . . . . . . X O X 4 8 |
$$ | . . . . . . . . . . . . . . . X X . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ -----------------------------------------[/go]

at
at

Click Here To Show Diagram Code

[go]$$Bcm11 White to play
$$ -----------------------------------------
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . O . . . O . O . . . . |
$$ | . . . . . . . . . . . . . . O X X 3 . |
$$ | . . . . . . . . . . . . . . X O X O O |
$$ | . . . . . . . . . . . . . . 1 O O 2 O |
$$ | . . . . . . . . . . . . . . X O X O O |
$$ | . . . . . . . . . . . . . . . X X 5 6 |
$$ | . . . . . . . . . . . . . . . . . 4 . |
$$ | . . . . . . . . . . . . . . . . 7 . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ -----------------------------------------[/go]

My question concerns only the last move
Is it the best move to capture the white stones?

Isn't it better to play

Click Here To Show Diagram Code

[go]$$Bcm11 White to play
$$ -----------------------------------------
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . O . . . O . O . . . . |
$$ | . . . . . . . . . . . . . . O X X X . |
$$ | . . . . . . . . . . . . . . X O X O O |
$$ | . . . . . . . . . . . . . . X O O O O |
$$ | . . . . . . . . . . . . . . X O X O O |
$$ | . . . . . . . . . . . . . . . X X X O |
$$ | . . . . . . . . . . . . . . . . 7 O . |
$$ | . . . . . . . . . . . . . . . . . 8 . |
$$ | . . . . . . . . . . . . . . . . 9 . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . a . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ -----------------------------------------[/go]

in order to avoid white sente move around "a"?

Remark
We have studied the following theorems.


Definitions
The following local endgames M0|F0 and M1|F1 each with one simple follow-up have the sente(?) move values M0, M1 and follow-up move values F0, F1.


Presuppositions
Suppose the local endgames {2*F0 | 0 || -M0} and {2*F1 | 0 || -M1} with M0 ≥ M1 and F0 ≥ F1 and a constant number C.

Theorem 1
Playing to {2*F0 | 0} + {2*F1 | 0 || -M1} is at least as good as playing to {2*F1 | 0} + {2*F0 | 0 || -M0}.


Definitions
If M0 ≥ M1 and F0 ≥ F1, we write M0|F0 for the larger local endgame and M1|F1 for the smaller local endgame.


Presuppositions
Suppose the local endgames with simple follow-ups M0|F0 and M1|F1 of the creator with M0 ≥ M1 and F0 ≥ F1 in a (possibly empty) environment E without kos now or later.

Theorem 2
The creator's start in M0|F0 is at least as good as in M1|F1.


Presuppositions
Suppose the local endgames each with one follow-up M0|F0 and M1|F1 of the creator with M0 ≥ M1 and M0 + 2*F0 ≥ M1 + 2*F1 in a (possibly empty) environment E without kos now or later.

Theorem 3
The preventer's start in M0|F0 is at least as good as in M1|F1.


Presuppositions
Suppose the local endgames each with one follow-up A|B and C|D of the creator with A ≥ C and A + 2B ≥ C + 2D.

Theorem 4
A correct start of the creator is playing at A.


Remarks and Questions
- For theorems 1 + 2, you have said that the local endgames can be local gotes, ambiguous or local sentes.
- For theorem 1, it is obvious that M0 and M1 are sente move values.
- For theorem 2, we have seen in the proof that {2*F0 | 0 || -M0} and {2*F1 | 0 || -M1} appear so we also know that M0 and M1 are sente move values.
- Is it correct for theorem 3 that M0 and M1 are sente move values even if one or both local endgames are local gotes?
- Is it correct for theorem 4 that A and C are sente move values even if one or both local endgames are local gotes?

RobertJasiek wrote:Remark
We have studied the following theorems.


Definitions
The following local endgames M0|F0 and M1|F1 each with one simple follow-up have the sente(?) move values M0, M1 and follow-up move values F0, F1.


Presuppositions
Suppose the local endgames {2*F0 | 0 || -M0} and {2*F1 | 0 || -M1} with M0 ≥ M1 and F0 ≥ F1 and a constant number C.

Theorem 1
Playing to {2*F0 | 0} + {2*F1 | 0 || -M1} is at least as good as playing to {2*F1 | 0} + {2*F0 | 0 || -M0}.


Definitions
If M0 ≥ M1 and F0 ≥ F1, we write M0|F0 for the larger local endgame and M1|F1 for the smaller local endgame.


Presuppositions
Suppose the local endgames with simple follow-ups M0|F0 and M1|F1 of the creator with M0 ≥ M1 and F0 ≥ F1 in a (possibly empty) environment E without kos now or later.

Theorem 2
The creator's start in M0|F0 is at least as good as in M1|F1.


Presuppositions
Suppose the local endgames each with one follow-up M0|F0 and M1|F1 of the creator with M0 ≥ M1 and M0 + 2*F0 ≥ M1 + 2*F1 in a (possibly empty) environment E without kos now or later.

Theorem 3
The preventer's start in M0|F0 is at least as good as in M1|F1.


Presuppositions
Suppose the local endgames each with one follow-up A|B and C|D of the creator with A ≥ C and A + 2B ≥ C + 2D.

Theorem 4
A correct start of the creator is playing at A.


Remarks and Questions
- For theorems 1 + 2, you have said that the local endgames can be local gotes, ambiguous or local sentes.
- For theorem 1, it is obvious that M0 and M1 are sente move values.
- For theorem 2, we have seen in the proof that {2*F0 | 0 || -M0} and {2*F1 | 0 || -M1} appear so we also know that M0 and M1 are sente move values.
- Is it correct for theorem 3 that M0 and M1 are sente move values even if one or both local endgames are local gotes?
- Is it correct for theorem 4 that A and C are sente move values even if one or both local endgames are local gotes?

IIRC, {2F0 | 0 || -M0} + C0 in CGT notation is equivalent to M0|F0 in your notation, where C0 is a number, but F0 and M0 need not be. If F0 and M0 are numbers, then we can use a difference game to prove things about M0|F0 and M1|F1. In that case, whether we call the positions sente or gote is irrelevant to the proof.

These days I have a lot to do, and I am ailing. Not with any serious ailment, but enough to make it difficult to get things done.

In a game {A||B|C} or {A|B||C}, M|F stands for the numbers that are the (hopefully sente) move value M and follow-up move value F. There, B need not equal 0 while your sente followers are 0.

Given two simple endgames in an environment with no ko caveat, G0 = {2*f0 | 0 || -m0} + c0 and G1 = {2*f1 | 0 || - m1} + c1, where f0, f1, m0, m1 are numbers ≥ 0 and c0 and c1 are numbers: WOLOG, let f0 ≥ f1.

1) If f0 = f1, it is better for Black to play in the game with the greater of m0 or m1.

2) If f0 > f1:
a) if m0 ≥ m1, it is better for Black to play in G0;
b) if m1 > m0 and 2*f0 > 2*f1 + m1, it is better for Black to play in G0;
c) if m1 > m0 and 2*f1 + m1 ≥ f0, which game is better for Black to play in depends upon the environment.

These rules do not depend upon local sente or gote, but when G0 is local sente, it may be right to play in G0, no matter how small m0 is. Thus, the usual heuristic of making the hotter play may be wrong.

I need to understand some basics of adding a number K (call it the komi, if you like) to a game.


{2F | 0 || -M} has the sente move value 0 - (-M) = M and follow-up move value (2F - 0) / 2 = F.

{2F + K | 0 + K || -M + K} has the sente move value (0 + K) - (-M + K) = M and follow-up move value ((2F + K) - (0 + K)) / 2 = F.

Is {2F | 0 || -M} + K = {2F + K | 0 + K || -M + K}?


Presuppositions
Suppose the local endgames {2*F0 | 0 || -M0} and {2*F1 | 0 || -M1} with M0 ≥ M1 and F0 ≥ F1 and a constant number C.

Theorem 1
Playing to {2*F0 | 0} + {2*F1 | 0 || -M1} is at least as good as playing to {2*F1 | 0} + {2*F0 | 0 || -M0}.


Hence, may we also write the following?

Presuppositions
Suppose the local endgames {2*F0 + K | 0 + K || -M0 + K} and {2*F1 + K | 0 + K || -M1 + K} with M0 ≥ M1 and F0 ≥ F1 and a constant number (C - K).

Theorem 1 + K
Playing to {2*F0 + K | 0 + K} + {2*F1 + K | 0 + K || -M1 + K} is at least as good as playing to {2*F1 + K | 0 + K} + {2*F0 + K | 0 + K || -M0 + K}.

RobertJasiek wrote:I need to understand some basics of adding a number K (call it the komi, if you like) to a game.

I follow the convention of using lower case letters for numbers and upper case letters for games (which might also be a number). In what follows below I will take all of your letters to indicate numbers, to keep things simple.

RobertJasiek wrote:Is {2F | 0 || -M} + K = {2F + K | 0 + K || -M + K}?

If F, M ≥ 0, yes.

Note: It is important that {2F | 0 || -M} is not a number.

RobertJasiek wrote:Presuppositions
Suppose the local endgames {2*F0 | 0 || -M0} and {2*F1 | 0 || -M1} with M0 ≥ M1 and F0 ≥ F1 and a constant number C.

Theorem 1
Playing to {2*F0 | 0} + {2*F1 | 0 || -M1} is at least as good as playing to {2*F1 | 0} + {2*F0 | 0 || -M0}.


Hence, may we also write the following?

Presuppositions
Suppose the local endgames {2*F0 + K | 0 + K || -M0 + K} and {2*F1 + K | 0 + K || -M1 + K} with M0 ≥ M1 and F0 ≥ F1 and a constant number (C - K).

Theorem 1 + K
Playing to {2*F0 + K | 0 + K} + {2*F1 + K | 0 + K || -M1 + K} is at least as good as playing to {2*F1 + K | 0 + K} + {2*F0 + K | 0 + K || -M0 + K}.

To compare moves in an environment without a ko caveat, we may set up a difference game.

Thus, given G0 = {2*f0 | 0 || - m0} + c0 and G1 = {2*f1 | 0 || - m1 + c1, environment E, and komi, k, we start by setting up this 0 game:

{2*f0 | 0 || - m0} + c0 + {2*f1 | 0 || - m1 + c1 + E + k +
{m0 || 0 | -2*f0} - c0 + (m1 || 0 | -2*f1} - c1 - E - k
=
{2*f0 | 0 || - m0} + {2*f1 | 0 || - m1 +
{m0 || 0 | -2*f0} + (m1 || 0 | -2*f1}

Note that c0, c1, E, and k all drop out.

(You have omitted three } brackets.)

***

F, M ≥ 0 because they are move values.

{2F | 0 || -M} must be interesting to play as an unsettled game so, for that purpose, I understand why it should not be a number. Do you have another reason in mind why it should not be a number?

***

I see all those extra numbers dropping out in your difference game but what do you want to tell me with it for "theorem 1 + K"?

***

Since {2F | 0 || -M} + K = {2F + K | 0 + K || -M + K}, let me discuss the following again:

Presuppositions
Suppose the local endgames with simple follow-ups M0|F0 and M1|F1 of the creator with M0 ≥ M1 and F0 ≥ F1 in a (possibly empty) environment E without kos now or later.

Theorem 2
The creator's start in M0|F0 is at least as good as in M1|F1.

Simply speaking, the proof works like this: [Strategy yields] the local endgames {2*F0 | 0 || -M0} and {2*F1 | 0 || -M1}, where we assume 0 at the sente followers as our calibration WLOG. Therefore, by theorem 1, theorem 2 is proved for its presuppositions. That is, the creator's two local endgames have arbitrary numbers of their sente followers so that the move values are M0|F0 and M1|F1, where M0 and M1 are sente move values indeed, even for local gote(s).

***

A closer reading of theorem 3 and its proof comes next to verify or refute _sente_ move values even for local gotes.